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Denote any number divisible by 3 \(D\) and any number not divisible by 3 \(N\) . The number range 100-999 starts with \(NNDNND\) ... and ends with ... \(NNDNND\) . In all, there are \(\frac{999 - 100 + 1}{3} = 300\) \(NNDs\) and thus there are \(2*300 = 600 Ns\) .

Would somebody please explain the +1 in the numerator of the given fraction? I don't quite understand why 1 is added.

What's the difference between your approach and the arithmetic progression? Here, both get the same result, but for the example you have in the basic section ("How many numbers betw. 12 and 96 div. by 4?") the arith. progr. approach doesn't work!? Thx.

The formula I wrote is basically the one counting # of terms in AP - \(# \ of \ terms=\frac{Last \ term -First \ term}{common \ difference}+1\) (\(n=\frac{a_n-a_1}{d}+1\) as \(a_n=a_1+d(n-1)\)), as multiples of some integers basically are AP.

So for the question: "How many multiples of 4 are there between 12 and 96, inclusive?" the formula will be the same and will give the same answer --> \(96=12+4(n-1)\) --> \(n=22\).

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20 Nov 2008, 21:08

snowy2009 wrote:

How many three-digit integers are not divisible by 3 ?

(C) 2008 GMAT Club - m16#11

* 599 * 600 * 601 * 602 * 603

Denote any number divisible by 3 \(D\) and any number not divisible by 3 \(N\) . The number range 100-999 starts with \(NNDNND\) ... and ends with ... \(NNDNND\) . In all, there are \(\frac{999 - 100 + 1}{3} = 300\) \(NNDs\) and thus there are \(2*300 = 600 Ns\) .

Would somebody please explain the +1 in the numerator of the given fraction? I don't quite understand why 1 is added.

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23 Jun 2010, 18:24

i think its worth mentioning also that when you do the arithmatic progression equation and divide by 3 or whatever number, it is not necessary that the number is wholly divisible in this case....just round down.,...

What's the difference between your approach and the arithmetic progression? Here, both get the same result, but for the example you have in the basic section ("How many numbers betw. 12 and 96 div. by 4?") the arith. progr. approach doesn't work!? Thx.

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22 Jun 2011, 17:04

I think B.3-digit integers 100, 101,102, ..., 999you can see that every 3 3-digit integers, there will be 1 3-digit integer that is divisible by 3 and number of 3-digit integer = 999-99 = 900As a result, number of 3-digit integers that is divisible by 3 = 900/3 = 300number of 3-digit integers that is NOT divisible by 3 = 900-300 =600
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